Introduction to Probability - Chapter 2 docx
Introduction to Probability - Chapter 2 docx
... to wait. Write a program to see if your conjecture is right.
2. 2. CONTINUOUS DENSITY FUNCTIONS 65
-1 -0 .5 0.5 1 1.5 2
0 .2
0.4
0.6
0.8
1
-1 -0 .5 0 0.5 1 1.5 2
0 .25
0.5
0.75
1
1 .25
1.5
1.75
2
F ... - z
1 - z
1 - z
1 - z
E
Figure 2. 19: Calculation of F
Z
.
20 40 60 80 100 120
0.005
0.01
0.015
0. 02
0. 025
0.03
f (t) = (1/30) e
- (1/30) t...
Introduction to Probability - Chapter 5 docx
... 28 92
7 3657 8 3 025 9 33 62
10 29 85 11 3138 12 3043
13 26 90 14 24 23 15 25 56
16 24 56 17 24 79 18 22 76
19 23 04 20 1971 21 25 43
22 26 78 23 27 29 24 24 14
25 26 16 26 24 26 27 23 81
28 20 59 29 20 39 30 22 98
31 ... DENSITIES
Democrat Republican
Female 24 4 28
Male 8 14 22
32 18 50
Table 5 .2: Observed data.
Democrat Republican
Female s
11
s
12
t...
Introduction to Probability - Chapter 8 docx
... 1
0
0. 02
0.04
0.06
0.08
0.1
0. 12
0.14
0 0 .2 0.4 0.6 0.8 1
0
0. 02
0.04
0.06
0.08
0.1
0. 12
0 0 .2 0.4 0.6 0.8 1
0
0.05
0.1
0.15
0 .2
0 .25
0 0 .2 0.4 0.6 0.8 1
0
0. 025
0.05
0.075
0.1
0. 125
0.15
0.175
n=10 n =20
n=40
n=30
n=60
n=100
Figure ... E(X
i
)=
1
0
xdx=
1
2
,
σ
2
= V (X
i
)=
1
0
x
2
dx −µ
2
=
1
3
−
1
4
=
1
12
.
Hence,
E
S
n
n
=
1
2
,
V
S
n
n
=
1...
Introduction to Probability - Chapter 1 pps
... DISCRETE PROBABILITIES 3
.20 3309 .7 620 57 .151 121 . 623 868
.9 320 52 .415178 .716719 .9674 12
.069664 .6709 82 .3 523 20 .049 723
.75 021 6 .784810 .089734 .966730
.946708 .380365 . 027 381 .900794
Table 1.1: ... PROBABILITIES 5
5
10 15 20 25 30 35 40
-1 0
-8
-6
-4
-2
2
4
6
8
10
Figure 1.1: Peter’s winnings in 40 plays of heads or tails.
One can understand this cal...
Introduction to Probability - Chapter 3 ppt
... ways to assign the first element, for
94 CHAPTER 3. COMBINATORICS
n = 0 1
10 1 10 45 120 21 0 25 2 21 0 120 45 10 1
9 1 9 36 84 126 126 84 36 9 1
8 1 8 28 56 70 56 28 8 1
7 1 7 21 35 35 21 7 ... {a
1
,a
2
,a
3
,a
4
} can be written in the
form
σ =
123 4
21 43
,
indicating that a
1
went to a
2
, a
2
to a
1
, a
3
to a
4
, and a
4
to a
3
.
Ifwealwayschoosetheto...
Introduction to Probability - Chapter 4 doc
...
156 CHAPTER 4. CONDITIONAL PROBABILITY
Y
-1 0 12
X -1 0 1/36 1/6 1/ 12
0
1/18 0 1/18 0
1
0 1/36 1/6 1/ 12
2
1/ 12 0 1/ 12 1/6
Table 4.6: Joint distribution.
36 A die is thrown twice. Let X
1
and X
2
denote ... Smith
Unconditional
probability
b
bb
b
g
g
g
b
1/4
1/8
1/8
1/8
1/8
1/4
1/4
1/4
1/4
1/4
1
1 /2
1 /2
1 /2
1 /2
1
bg
gb
gg
b
bb
b
b
1/4
1/8
1/8
1/4
1/4
1/4
1
1 /2...
Introduction to Probability - Chapter 6 doc
... Y )
2
) −(a + b)
2
= E(X
2
)+2E(XY )+E(Y
2
) −a
2
− 2ab − b
2
.
Since X and Y are independent, E(XY )=E(X)E(Y )=ab. Thus,
V (X + Y )=E(X
2
) −a
2
+ E(Y
2
) −b
2
= V (X)+V (Y ) .
✷
24 6 CHAPTER ... value 0:
p
X
=
2 −10 12
3/11 2/ 11 1/11 2/ 11 3/11
;
p
Y
=
2 −10 12
1/11 2/ 11 5/11 2/ 11 1/11
.
25 6 CHAPTER 6. EXPECTED VALUE AND VARIANCE
37 The read...
Introduction to Probability - Chapter 7 pptx
... f
Z
if
f
X
(x)=
1
√
2 σ
1
e
−(x−µ
1
)
2
/2
2
1
f
Y
(x)=
1
√
2 σ
2
e
−(x−µ
2
)
2
/2
2
2
.
*7 Suppose that R
2
= X
2
+ Y
2
. Find f
R
2
and f
R
if
f
X
(x)=
1
√
2 σ
1
e
−(x−µ
1
)
2
/2
2
1
f
Y
(x)=
1
√
2 σ
2
e
−(x−µ
2
)
2
/2
2
2
.
8 ... with λ =1 /2, β =1 /2 (see Example 7.4). Now let
R
2
= X
2
+ Y
2
. Then
f
R
2
(r)=
+∞
−∞
f
X
2
(r − s)f
Y
2
(s)...
Introduction to Probability - Chapter 9 pps
... E(X)=7 /2 and σ
2
= V (X)=35/ 12. Thus, E(S
420
)=
420 ·7 /2 = 1470, σ
2
(S
420
) = 420 ·35/ 12 = 122 5, and σ(S
420
) = 35. Therefore,
P (1400 ≤ S
420
≤ 1550) ≈ P
1399.5 −1470
35
≤ S
∗
24 0
≤
1550.5 ... .0000 1.0 .3413 2. 0 .47 72 3.0 .4987
.1 .0398 1.1 .3643 2. 1 .4 821 3.1 .4990
.2 .0793 1 .2 .3849 2. 2 .4861 3 .2 .4993
.3 .1179 1.3 .40 32 2.3 .4893 3.3 .4995...
Introduction to Probability - Chapter 10 pptx
...
10 .2. BRANCHING PROCESSES 387
Geometric
p
j
Data Model
0 .20 92 .1816
1 .25 84 .3666
2 .23 60 .20 28
3 .1593 .1 122
4 .0 828 .0 621
5 .0357 .0344
6 .0133 .0190
7 .00 42 .0105
8 .0011 .0058
9 .00 02 .00 32
10 ... FUNCTIONS
Z
1
Z
2
Z
3
Z
4
Z
5
Z
6
Z
7
Z
8
Z
9
Z
10
Z
11
Z
12
Profit
126 778119766 520 0
100000000000 -5 0
100000000000 -5 0
111000000000 -5 0
00000000000 0-1...
Từ khóa:
- introduction to network simulator 2 ns2
- introduction to the java 2 micro edition platform api
- introduction to probability theory and statistics
- introduction to asp net 2 0
- introduction to probability theory and statistical analysis
- introduction to probability theory
- introduction to this chapter
- chapter 2 introduction to data structures and
- chapter 1 2 introduction to databases pdf